217s418sol.pdf - Exam 2 Solutions STAT/MATH 418 Spring 2017...

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Exam 2 Solutions STAT/MATH 418 Spring 2017 1. As the mean of the exponential is given to be 5 minutes, the rate parameter λ = 1 / 5. The lack of memory property of the exponential thus gives the result e - 4 / 5 . 2. 1 = R 1 0 R 5 1 c y dy dx = R 1 0 12 c dx = 12 c. Hence c = 1 / 12. X and Y are independent as the joint density f X,Y can be expressed as product of functions g and h , where g ( x ) = 1 12 for 0 < x < 1 and zero otherwise, and h ( y ) = y for 1 < y < 5 and zero otherwise. 3. f X ( x ) = 1 if 0 < x < 1 and 0 otherwise. Let g ( x ) = - 2 log x . y = - 2 log x implies x = e - y/ 2 , so g - 1 ( y ) = e - y/ 2 for y > 0. Use the formula f Y ( y ) = f X ( g - 1 ( y )) | d dy g - 1 ( y ) | if y = g ( x ) for some x and 0 otherwise, to get f Y ( y ) = 1 2 e - y/ 2 if y > 0 and zero otherwise. So Y is distributed as exponential with λ = 1 2 . 4. Let X denote distance from A to the breakdown site, and let L denote the distance from the breakdown site to the nearest service station. So, L = 25 - X, if 0 < X < 25 X - 25 , if 25 X < 37 . 5 | X - 50 | if 37 . 5 X < 62 . 5 75 - X if 62 . 5 X < 75 X - 75 , if 75 X 100 . E ( L ) = 2 100 Z 25 0 xdx + 4 100 Z 12 . 5 0 xdx = 25 4 + 12 .
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